$S^2 \times S^2$ is diffeomophic to $G_2(\mathbb{R}^4)$

Question

$G_2(\mathbb{R}^4)$ is the Grassmannian manifold of two-dimensional subspaces of $\mathbb{R}^4$. I would like a detailed proof. Can it be done explicitly? I mean, showing the map and checking its differentiability using charts?

Answer 1

I think something along the lines of what follows should work. I haven't yet checked smoothness (but everything is smooth), and there is a problem towards the end that prevents me from reaching a conclusion.

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Consider the Stiefel manifold $V_2(\Bbb R^4)$ of orthonormal $2$ frames, i.e. of pairs of vectors $(x,y)\in\Bbb R^4$such that $x\cdot y=0$ and $|x|=|y|=1$. The group $O(2)$ of real orthogonal $2\times 2$ matrices acts on $V_2(\Bbb R^4)$ by $$\begin{pmatrix} a&b\\c&d\end{pmatrix}\cdot(x,y)=(ax+by,cx+dy)$$ and the orbit of a point $(x,y)\in V_2(\Bbb R^4)$ under this action is the collection of all orthonormal bases of $\operatorname{Vect}(x,y)=\Bbb R x\oplus \Bbb R y$. Thus the canonical map $$\begin{align}\operatorname{Vect}:&V_2(\Bbb R^4)\to G_2(\Bbb R^4)\\ & (x,y)\mapsto\operatorname{Vect}(x,y)\end{align}$$ of which smoothness would have to be checked (which it is), gives rise to a diffeomorphism $$O(2)\backslash V_2(\Bbb R^4)\xrightarrow{~\sim~}G_2(\Bbb R^4)$$ On the other hand, let us consider $\Bbb R^4=\Bbb C^2$ through $(x_1,x_2,x_3,x_4)\mapsto(x_1+ix_2,x_3+ix_4)$ and the map $$\begin{align} \phi :&V_2(\Bbb R^4)\to\mathrm{P}^1(\Bbb C)\times\mathrm{P}^1(\Bbb C)\\&(x,x')\mapsto (\Bbb C x,\Bbb Cx')\end{align}$$ The complex projective line $\mathrm{P}^1(\Bbb C)$ is diffeomorphic to the two sphere $S^2$, so we may consider that $\phi$ maps to $S^2\times S^2$. Let us show that this map is onto. Let $l=\Bbb C x$ and $l'$ be two complex lines $\subset\Bbb C^2$ (i.e. $\in\mathrm{P}^1(\Bbb C)$). Then $l'\cap x^{\perp}\neq 0$, since $l'$ has real dimension $2$ and $x^{\perp}$ has real dimension $3$, and everything happens inside $\Bbb C^2$ which has real dimension $4$. Let $x'\in l'\cap x^{\perp}$ be of unit length, then $l'=\Bbb Cx'$ and $(l,l')=\phi(x,x')$.

Now let us solve (for $(x,x')\in V_2(\Bbb R^4)$) the equation $\phi(x,x')=(l,l')=\phi(x_0,x'_0)$. If $(x,x')\in V_2(\Bbb R^4)$ satisfy $\phi(x,x')=\phi(x_0,x'_0)$, then there exists $z=e^{i\theta},z'=e^{i\theta'}\in S^1$ (i.e. complex numbers of module one) with $x=zx_0$ and $x'=z'x'_0$, but we must also have $x\cdot x'=0(=x_0\cdot x'_0)$.

For $x_0=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\in\Bbb R^4\longleftrightarrow\begin{pmatrix}X_1\\X_2\end{pmatrix}\in\Bbb C^2$ and $x'_0=\begin{pmatrix}x_1'\\x_2'\\x_3'\\x_4'\end{pmatrix}\in\Bbb R^4\longleftrightarrow\begin{pmatrix}X_1'\\X_2'\end{pmatrix}\in\Bbb C^2$, $x=zx_0=\begin{pmatrix}zX_1\\zX_2\end{pmatrix}$, $x'=z'x'_0=\begin{pmatrix}z'X_1'\\z'X_2'\end{pmatrix}$ and $0=x_0\cdot x'_0=\mathrm{Re}(X_1\overline{X'_1})+\mathrm{Re}(X_2\overline{X'_2})=\mathrm{Re}(X_1\overline{X'_1}+X_2\overline{X'_2})$ while $$x\cdot x'=\mathrm{Re}(zX_1\overline{z'X'_1})+\mathrm{Re}(zX_2\overline{z'X'_2})=\mathrm{Re}\Big(e^{i(\theta-\theta')}(X_1\overline{X'_1}+X_2\overline{X'_2})\Big)$$

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Here's where I run into a problem, because I'd like to conclude that $e^{i(\theta-\theta')}=\pm 1$, but for that I'd need $X_1\overline{X'_1}+X_2\overline{X'_2}=\left\langle\begin{pmatrix}X_1\\X_2\end{pmatrix}|\begin{pmatrix}X_1'\\X_2'\end{pmatrix}\right\rangle\neq 0$, i.e. $l$ not orthogonal to $l'$ in the hermitian space $\Bbb C^2$, which would force $z=\pm z'$ and would give a left action of $O(2)$ on $V_2(\Bbb R^4)$ whose orbits are exactly the fibers of $\phi$. This would give a diffeomorphism $$O(2)\backslash V_2(\Bbb R^4)\xrightarrow{~\sim~}\mathrm{P}^1(\Bbb C)\times\mathrm{P}^1(\Bbb C)$$ Since $\mathrm{P}^1(\Bbb C)=S^2$ this would almost be enough, at least once we show that the two actions of $O(2)$ are conjugate.

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To add a little more detail on the action of $O(2)$ on $V_2(\Bbb R^4=\Bbb C^2)$, I consider at the end, it is obtained from $O(2)=S^1\times\lbrace -1,+1\rbrace$ and $$(z,\pm 1)\cdot (x,x')=(zx,\pm zx')\;.$$